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Thank you for the suggestion. I will do that.

Let n and k be two natural numbers and let S be a set of n points such that
(a) no three points of S are collinear.
(b) for any point P of S there are at least k points of S which are equidistant from P.
Prove that k<1/2+(2n)^(1/2)

Prove that for all positive integers n there is an n-digit multiple of 5^n all the the digits of which is odd.

Thanks to Anindya Biswas for solving a general form of the problem

Let X be the intersection of AD and BC. Then XA/XB=XD/XC =(XA-XD)/(XB-XC)=AD/BC=70/50=7/5. Then P lies on the bisector of <AXB. Therefore P is the intersection of side AB and the bisector of its opposite angle. So AP/PB=XA/XB=7/5. So AP=7AB/(7+5)=7×92/12=161/3.

So the answer is 161+3=164

Let A be the set of all such paths and let B be the set of all sequence of coordinates (x,y) such that both x and y are primes , if (x1,y1) is an entry of the sequence then then the next entry is (x2,y2) where x1=x2 and y2 is the smallest prime greater than y2 or vice versa and the first entry of th...